# How do you use a Taylor series to find the derivative of a function?

Sep 27, 2014

One of the benefits of a Taylor series is the ease of differentiation since it can be done term by term. So, the Talyor series

f(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}(x-a)^n

can be differentiated as

f'(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}[(x-a)^{n}]'

by Power Rule,

=sum_{n=1}^infty{f^{(n)}(a)}/{n!}[n(x-a)^{n-1}]

(Note: $n$ starts from $1$ since the term is zero when $n = 0$)
by simplifying a little further,

=sum_{n=1}^infty{f^{(n)}(a)}/{(n-1)!}(x-a)^{n-1}

I hope that this was helpful.